\documentclass[final]{amsart}
\usepackage{fly,cancel}
\usepackage{amsmath,amscd}
\usepackage{sidecap}
\title{Notes on Lie Groups and Algebras}
\DeclareMathOperator{\lie}{Lie}
\date{July 27, 2008}
\begin{document}
\maketitle
\tableofcontents
\section{Introduction}

We follow an approach similar to Artin~\cite{ArtinBk01}.

\input{tangent.tex}
\input{infinitesimal.tex}

\section{Tangent Vectors and Infinitesimals Together}

Recall Lemma (\ref{lemma1}) that tangent vectors are orthogonal to $\nabla f(x)$.
Given a point $x$ of $\mathbb{R}^n$ and a vector $\vec{v}\in\mathbb{R}^n$, the
sum $x + \vec{v}\varepsilon$ is a vector with infinitesimal entries which we
think of intuitively as \emph{an infinitesimal change in $x$ in the direction of
$\vec{v}$.} Observe the Taylor expansion of $f(x + \vec{v}\varepsilon)$ yields by
Eq (\ref{classicTaylor})
\begin{equation}
f(x + \vec{v}\varepsilon) = f(x) + \varepsilon\vec{v}\cdot\nabla f(x)
\end{equation}
since higher order terms vanish by our use of infinitesimal generator $\varepsilon$.

We want to find tangent vectors to some point $x\in\mathbb{R}^n$, that is vectors such that
\begin{equation}
\vec{v}\cdot\nabla f(x) = 0.
\end{equation}
If we demand that $x\in S$ is part of the algebraic set, then we necessarily
have
\begin{equation}\label{algebraicSet}
f(x) = 0
\end{equation}
since we specifically defined algebraic sets to be the sets of zeros of 
polynomials.

If we notice these two things, we have
\begin{equation}
f(x + \vec{v}\varepsilon) = f(x) + \varepsilon\vec{v}\cdot\nabla f(x) = \varepsilon\vec{v}\cdot\nabla f(x)
\end{equation}
since by Eq (\ref{algebraicSet}) we have $f(x)=0$ we are left with the right
hand side. Since we want to find all such vectors $\vec{v}$ that are orthogonal
to the gradient of $f$, we then can find it through the relationship
\begin{equation}
f(x + \vec{v}\varepsilon) = 0
\end{equation}
instead of solving a differential system of equations. We will be using this 
property a lot to find tangent vectors.

Here we must reiterate the beauty of this system. To find tangent vectors, we 
simply are looking for vectors $\vec{v}$ such that
\begin{equation}
f(x + \varepsilon\vec{v}) = 0
\end{equation}
where $f(x)=0$ generates an algebraic set $S$. By taylor expanding, we see that
\begin{equation}
f(x + \varepsilon\vec{v}) = \cancelto{0}{f(x)} + \varepsilon\vec{v}\cdot\nabla f(x)
\end{equation}
and moreover, if this is zero then we have found the tangent vectors to the
point $x\in S$!

\input{matrixgrps}
\input{lieAlgebra}

\appendix
\input{groups}
\input{matrix}
\bibliographystyle{utcaps}
\bibliography{liebib}
\end{document}
